三條有關三角比的數學

Q21,
Distance from top of wall to water level
= 5 x cos 40
= 3.83m

After 1 hour
Distance from top of wall to water level
= 3.83 - 1
= 2.83m

The new angle
= cos^(-1) [2.83 / 5]
= 55.53 degree
===========================
Q22,
(a)
Distance OA
= Speed x Time
= 30 x 3
= 90km

Distance of OB, In triangle AOB
= OA cos 60
= 45km

Speed of ship B
= Distance / Time
= 45 / 3
= 15km/h

(b)
Distance of AB, In triangle AOB
= OB tan 60
= 79.44km

Let t be the time taken for both ship met the other from 3pm
Since, AP + PB = AB
=> Speed A x t + Speed B x t = AB
=> t = AB/( Speed A + Speed B )
=> t = 79.44/(30+15)
=> t = 1.73h

Therefore,
PB
= Speed B x t
= 15 x 1.73
= 25.98km

In triangle POB
Angle POB
= tan^(-1) [PB / OB]
= 30 degree

In triangle APB
Angle of POA
= Angle of AOB - Angle of POB
= 60 - 30
= 30 degree
==========================
Q4,
**唔好意思我唔識用那角度符號, angle CBD 用 H 代表; angle CAD則用 K 代表**
(a)
Tan H
= CD/BD
= CD/3AD
= (CD/AD)/3
= (Tan K)/3

(bi)
AD
= AC x cos K
= 3 x cos 49
= 1.968

In triangle ACD,
CD = 3 x sin K
= 3 x sin 49
= 2.264

Since BD = 3AD
BD
= 3AD
= 5.905

In triangle BCD,
Angle H
= tan^(-1) [CD / BD]
= tan^(-1) [2.264 / 5.905]
= 21 degree (corr to 2 sig.fig)

(bii)
In triangle BCD
BC
= CD / sin H
= 2.264 / sin H
= 6.324

perimeter of triangle ABC
= AC + AD + BD + BC
= 3 + 1.968 + 5.905 + 6.324
= 17 units (corr to 2 sig.fig)

21.original distance between water level and pier =5cos40
new distance between water level and pier = 5cos40 - 1
let the required angel be x
cos x = (5cos40 - 1 ) / 5

22. AO = 30 x3 = 90km
BO = 90 cos 60
Speed = BO / time = 90 cos 60 / 3
第三張圖未睇到